Radicals as Inverse Polynomial Functions
Learning Objectives
- Verify that a radical and a polynomial function are inverses of each other
- Find the inverse of a polynomial function
[latex]\begin{array}{l} 18=a{6}^{2}\hfill \\ a=\frac{18}{36}\hfill \\ \text{ }=\frac{1}{2}\hfill \end{array}[/latex]
Our parabolic cross section has the equation[latex]y\left(x\right)=\frac{1}{2}{x}^{2}[/latex]
We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y the width will be given by 2x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative. To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable:[latex]\begin{array}{l}y=\frac{1}{2}{x}^{2}\hfill \\ 2y={x}^{2}\hfill \\ \text{ }x=\pm \sqrt{2y}\hfill \end{array}[/latex]
This is not a function as written. We are limiting ourselves to positive x values, so we eliminate the negative solution, giving us the inverse function we’re looking for.[latex]y=\frac{{x}^{2}}{2},\text{ }x>0[/latex]
Because x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. The trough is 3 feet (36 inches) long, so the surface area will then be:[latex]\begin{array}{l}\text{Area} & =l\cdot w\hfill \\ \text{ } & =36\cdot 2x\hfill \\ \text{ } & =72x\hfill \\ \text{ } & =72\sqrt{2y}\hfill \end{array}[/latex]
This example illustrates two important points:- When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
- The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.
[latex]{f}^{-1}\left(f\left(x\right)\right)=x,\text{for all }x\text{ in the domain of }f[/latex]
and[latex]f\left({f}^{-1}\left(x\right)\right)=x,\text{for all }x\text{ in the domain of }{f}^{-1}[/latex]
A General Note: Verifying Two Functions Are Inverses of One Another
Two functions, f and g, are inverses of one another if for all x in the domain of f and g. [latex-display]g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)=x[/latex-display]How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.
- Replace [latex]f\left(x\right)[/latex] with y.
- Interchange x and y.
- Solve for y, and rename the function [latex]{f}^{-1}\left(x\right)[/latex].
Example: Verifying Inverse Functions
Show that [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses, for [latex]x\ne 0,-1[/latex] .Answer: We must show that [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex].
[latex]\begin{array}{l}{f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(\frac{1}{x+1}\right)\hfill \\ \text{ }=\frac{1}{\frac{1}{x+1}}-1\hfill \\ \text{ }=\left(x+1\right)-1\hfill \\ \text{ }=x\hfill \\ f\left({f}^{-1}\left(x\right)\right)=f\left(\frac{1}{x}-1\right)\hfill \\ \text{ }=\frac{1}{\left(\frac{1}{x}-1\right)+1}\hfill \\ \text{ }=\frac{1}{\frac{1}{x}}\hfill \\ \text{ }=x\hfill \end{array}[/latex]
Therefore, [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses.Try It
Show that [latex]f\left(x\right)=\frac{x+5}{3}[/latex] and [latex]{f}^{-1}\left(x\right)=3x - 5[/latex] are inverses.Answer: [latex]{f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(\frac{x+5}{3}\right)=3\left(\frac{x+5}{3}\right)-5=\left(x - 5\right)+5=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=f\left(3x - 5\right)=\frac{\left(3x - 5\right)+5}{3}=\frac{3x}{3}=x[/latex]
Example: Finding the Inverse of a Cubic Function
Find the inverse of the function [latex]f\left(x\right)=5{x}^{3}+1[/latex].Answer: This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x.
[latex]\begin{array}{l}\text{ }y=5{x}^{3}+1\hfill \\ \text{ }x=5{y}^{3}+1\hfill \\ \text{ }x - 1=5{y}^{3}\hfill \\ \text{ }\frac{x - 1}{5}={y}^{3}\hfill \\ {f}^{-1}\left(x\right)=\sqrt[3]{\frac{x - 1}{5}}\hfill \end{array}[/latex]
Analysis of the Solution
Look at the graph of f and [latex]{f}^{-1}[/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[/latex]. This is always the case when graphing a function and its inverse function. Also, since the method involved interchanging x and y, notice corresponding points. If [latex]\left(a,b\right)[/latex] is on the graph of f, then [latex]\left(b,a\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Since [latex]\left(0,1\right)[/latex] is on the graph of f, then [latex]\left(1,0\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Similarly, since [latex]\left(1,6\right)[/latex] is on the graph of f, then [latex]\left(6,1\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex].Try It
Find the inverse function of [latex]f\left(x\right)=\sqrt[3]{x+4}[/latex].Answer: [latex]{f}^{-1}\left(x\right)={x}^{3}-4[/latex]
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- Question ID 15856. Authored by: Sousa,James. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].